The line whose vector equation are `r=2hat(i)-3hat(j)+7hat(k)+lambda(2hat(i)+phat(j)+5hat(k)) and r=hat(i)+2hat(j)+3hat(k)+mu(3hat(i)-phat(j)+phat(k))` are perpendicular for all values of `lambda and mu ` if p eqauls to

Options

A. `-1`
B. `2`
C. `5`
D. `6`

Answer

To solve the problem, we need to determine the value of \( p \) such that the two given lines are perpendicular for all values of \( \lambda \) and \( \mu \).

### Step-by-Step Solution:

1. **Identify the direction ratios of the lines**:
The vector equations of the lines are given as:
\[ \mathbf{r_1} = 2\hat{i} - 3\hat{j} + 7\hat{k} + \lambda(2\hat{i} + p\hat{j} + 5\hat{k}) \]
\[ \mathbf{r_2} = \hat{i} + 2\hat{j} + 3\hat{k} + \mu(3\hat{i} - p\hat{j} + p\hat{k}) \]
From these equations, we can extract the direction ratios:
- For line 1, the direction vector \( \mathbf{b_1} = 2\hat{i} + p\hat{j} + 5\hat{k} \)
- For line 2, the direction vector \( \mathbf{b_2} = 3\hat{i} - p\hat{j} + p\hat{k} \)

2. **Set up the condition for perpendicularity**:
Two lines are perpendicular if the dot product of their direction vectors is zero:
\[ \mathbf{b_1} \cdot \mathbf{b_2} = 0 \]
Therefore, we compute:
\[ (2\hat{i} + p\hat{j} + 5\hat{k}) \cdot (3\hat{i} - p\hat{j} + p\hat{k}) = 0 \]

3. **Calculate the dot product**:
Expanding the dot product:
\[ 2 \cdot 3 + p \cdot (-p) + 5 \cdot p = 0 \]
This simplifies to:
\[ 6 - p^2 + 5p = 0 \]

4. **Rearrange the equation**:
Rearranging gives us:
\[ -p^2 + 5p + 6 = 0 \]
Multiplying through by -1:
\[ p^2 - 5p - 6 = 0 \]

5. **Factor the quadratic equation**:
We can factor this quadratic:
\[ (p - 6)(p + 1) = 0 \]

6. **Solve for \( p \)**:
Setting each factor to zero gives us:
\[ p - 6 = 0 \quad \Rightarrow \quad p = 6 \]
\[ p + 1 = 0 \quad \Rightarrow \quad p = -1 \]

Thus, the values of \( p \) for which the lines are perpendicular are \( p = 6 \) and \( p = -1 \).

### Final Answer:
The values of \( p \) are \( p = 6 \) or \( p = -1 \).